The magnetic flux that passes through one turn of a 8-turn coil of wire changes to 5.0 Wb from 8.0 Wb in a time of 0.098 s. The average induced current in the coil is 140 A. What is the resistance of the wire

Answers

Answer 1

Answer:

Resistance is 1.75 ohms

Explanation:

Magnetic flux:

[tex]{ \phi{ = NBA}}[/tex]

N is number of turns, N = 8

B is magnetic flux

A is area of projection.

From faradays law:

[tex]E = - \frac{ \triangle \phi}{t} [/tex]

where E is the Electro motive force.

But E = IR

where I is current and R is resistance:

[tex]IR = \frac{( \phi_{1} - \phi _{2}) }{t} \\ \\ 140 \times R = \frac{8 \times (8 - 5)}{0.098} \\ \\ R = \frac{24}{0.098 \times 140} \\ \\ resistance = 1.75 \: ohms[/tex]


Related Questions

a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.​

Answers

Explanation:

It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is

[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]

[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]

a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force​

Answers

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?

Answers

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]

Speed of rocket 1 with respect to rocket 2 :

[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]

[tex]$u' = \frac{0.7 c}{1.12}$[/tex]

[tex]u'=0.625 c[/tex]

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.

Answers

Answer:

the body is subjected to a continuous rotation and the body is not in rotational equilibrium

Explanation:

For an object to have a static equilibrium, it must meet two relationships

             ∑ F = 0

             ∑ τ =0              

force acting on a body fulfills the relation of

         sum F = F - F = 0

when two forces do not move from position.

To find the torque we assume that the counterclockwise rotations are positive

        Σ τ = - F r - F r

        Στ = -2 Fr <> 0

consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium

The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.

Answers

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at a given distance:___________.
A) must be directed inward.
B) acts like it originated in a point charge Q at the center of the distribution.
C) must be directed outward.
D) must be greater than zero.
E) must be zero.

Answers

Answer:

Q at the center of the distribution.

Explanation:

The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.

which of the following cannot be increased by using a machine of some kind? work, force, speed, torque

Answers

Explanation:

Work cannot be increased by using a machine of some kind.

Work cannot be increased by using a machine of some kind.

A machine is any device in which the effort applied at one end overcomes a load at the other end.

Machines are generally used to perform different tasks faster.

However, a simple machine can not be used to increase the amount of work done at any time.

Force, speed and torque can all be increased using machines.

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Which of the following groups is the largest ?

population
community
ecosystem
biome

Answers

Answer:

B. Community

Took science classes for 6 years now

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. Group of answer choices

Answers

Answer:

check photo for solve

Explanation:

State the law of conservation of momentum

Answers

Explanation:

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant

Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.

Answers

D. I hope my answer helps you!

A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.

a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?

b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?

Answers

Answer: it would be daddy

Explanation:

Because I’m daddy

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars

Answers

Answer:

the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Explanation:

Given the data in the question;

Time period = 6 months = 1.577 × 10⁷ s

orbital speed v = 64000 m/s

since its a circular orbit,

v = 2πr / T

we solve for r

r = vT/ 2π

r = ( 64000 × 1.577 × 10⁷ ) / 2π

r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU

Now, from Kepler's law

T² = r³ / ( m₁ + m₂ )

T = 6 months = 0.5 years

we substitute

(0.5)² = (1.0737)³ / ( m₁ + m₂ )

0.25 = 1.2378 / ( m₁ + m₂ )

( m₁ + m₂ ) = 1.2378 / 0.25

( m₁ + m₂ ) = 4.9512

m₁ = m₂  = 4.9512 / 2 = 2.4756 solar mass

we know that solar mass = 1.989 × 10³⁰ kg

so

m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg

m₁ = m₂ = 4.92 × 10³⁰ kg

Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection

Answers

Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction

Answers

Answer:

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Explanation:

The expression for the angle of refraction is

          n₁ sin θ₁ = n₂ sin θ₂

refractive index n₁ is for incident light and n₂ is for transmitted light.

We have two cases

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.

Answers

Answer:

I think D) less than 50 years

Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.

Who is an astronaut?

An astronaut observes and performs the experiments based on the universe.

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.

Due to special relativity, between space and Earth, both moving with different speeds.

The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.

So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.

Thus, the correct option is d.

Learn more about astronaut.

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Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​

Answers

[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

In which situation should a parent be proactive and act to assume responsibility?

Answers

Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.

Explanation: i picked this and it is correct, you’re welcome:)

Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?

Answers

Answer:

The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.

Explanation:

Consider newtonian mechanics here.

Dynamic equation is

The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.

we'll consider newtonian mechanics here.

so the dynamic equations is S = ut + 0.5at^2

we know u=0.8

S=1.2+0.9

t=1.2/1.5

find a.

What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making

Answers

Hi! I believe the answer is C. or The seasons because all the other options use springs when making cars, bikes, and clocks. I hope this helped. Goodluck :)

3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic

Answers

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

Learn more about refractive index of light here;

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What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm

Answers

Hi! The answer is 15km I'm pretty sure

Answer:

The submarine's maximum safe depth in sea water is 801.678 m.

Explanation:

P=Po+(rho)*g*h

Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)

Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2

Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa

Initial Pressure= 1atm= 101.3kPa= 101300Pa

Water Density (rho) = 1000kg/m^3

Gravity= 9.8m/s^2

So rearranging for h= (P-Po)/((rho)*g)

h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m

Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed​

Answers

Answer:

c.

magnitude of displacement

A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.

Answers

Answer:

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium

[tex]d_0[/tex] is the object distance

[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium

[tex]d_i[/tex] is the image distance

R is the radius of curvature

Now, let [tex]d_0[/tex] = ∞, such that;

[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

we make [tex]d_i[/tex] subject of the formula

[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )

[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )

given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00

so we substitute

[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

[tex]d_i[/tex] = 9.165 / 0.41

[tex]d_i[/tex] = 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm

15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?

Answers

Answer: (15 - 0)/1.8 = 8. 33m/s^2

Explanation:

The acceleration of the racehorse is 8.33 m/s²

The given parameters;

initial velocity of the racehorse, u = 0

final velocity of the racehorse, v = 15  m/s

time of motion of the horse, t = 1.8 s

The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;

[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]

Thus, the acceleration of the racehorse is 8.33 m/s²

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Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A. If they are separated by a distance 4.3 m, at what location between the wires is the net magnetic field be twice the strength of the magnetic field from wire 1

Answers

Answer:

x= 2*I1*d/(I1+I2) meter

Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for longer wavelengths of light. The work function of the surface is

Answers

Explanation:

Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]

[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]

[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]

The work function [tex]\phi[/tex] is then

[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]

[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]

The work function of the surface is equal to 3.98 × 10⁻¹⁹J.

What are frequency and wavelength?

The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.

The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.

The expression between wavelength (λ), frequency, and speed of light (c) is:

c = νλ

Given, the wavelength of the light, ν = 500 nm

The frequency of the light can determine from the above-mentioned relationship:

ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz

The work function = h ν =  6 × 10¹⁴ × 6.626 × 10⁻³⁴

φ = 3.98 × 10⁻¹⁹J

Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.

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Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]​

Answers

Answer:

Gap left = Change in length on heating

Gap=Initial length×Coefficient of linear expansion×change in temperature

Gap=10×0.000012×15m

⟹Gap=0.0018 m

this is an example u have to put your equation in it

A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.

Answers

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event

Answers

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

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